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        general >> complex analysis >> power of cosine function
        
(Message started by: comehome1981 on Oct 25^th, 2010, 8:40pm)

Title: power of cosine function
Post by comehome1981 on Oct 25^th, 2010, 8:40pm

A question as follows:

It is clear that

[cos(k/n)]^(n^2) ----> exp{-k^2/2} as n goes to infinity

Does this hold when k=n/2 or n or some fraction of n?

Another question:

Does one know the estimate of cos(x) as x --> pi/2

Thanks for any tips!

Title: Re: power of cosine function
Post by towr on Oct 26^th, 2010, 1:17am

on 10/25/10 at 20:40:25, comehome1981 wrote:

It is clear that

[cos(k/n)]^(n^2) ----> exp{-k^2} as n goes to infinity

I've tried graphing it for a few k, and I must say, it's anything but clear. [cos(k/n)]^(n^2) / exp{-k^2} doesn't seem to convergence on 1.

[edit]Ah, wolframalpha says it should be exp(- 1/2 k^2)[/edit]

Quote:

Does one know the estimate of cos(x) as x --> pi/2

Isn't it just zero? cos(pi/2)=0

Title: Re: power of cosine function
Post by pex on Oct 26^th, 2010, 1:32am

on 10/26/10 at 01:17:10, towr wrote:

Ah, wolframalpha says it should be exp(- 1/2 k^2)

Taking logs and using L'Hopital twice also gives this result ;)

Title: Re: power of cosine function
Post by pex on Oct 26^th, 2010, 1:36am

on 10/25/10 at 20:40:25, comehome1981 wrote:

Does this hold when k=n/2 or n or some fraction of n?

No, it doesn't. If k = cn, clearly k/n = c and you're taking the limit of [cos(c)]^{n^2}. This limit is obviously either 1 (if cos(c)=1), nonexistent (if cos(c)=-1), or zero (otherwise).

Title: Re: power of cosine function
Post by towr on Oct 26^th, 2010, 5:20am

on 10/26/10 at 01:36:29, pex wrote:

No, it doesn't. If k = cn, clearly k/n = c and you're taking the limit of [cos(c)]^{n^2}. This limit is obviously either 1 (if cos(c)=1), nonexistent (if cos(c)=-1), or zero (otherwise).

But the limit of exp{- 1/2 k^2} = exp{- 1/2 (cn)^2} would also be 0. So the two expressions might still converge. (Were it not that as far as I can tell they generally don't.)

[edit]
If you want
[cos(c)]^{n^2} -> exp{- 1/2 (cn)^2}
then you must have
cos(c) -> exp{- 1/2 c^2}
Since c is constant, the expressions on both sides have to be equal, and thus c must be 0. (Which means it falls under the original case, since k=cn is constant for c=0.)
[/edit]

Title: Re: power of cosine function
Post by comehome1981 on Oct 26^th, 2010, 12:24pm

For the second part, I meant

is there a formula for the error bound for
|cos(x)| when x closes to pi/2

Title: Re: power of cosine function
Post by towr on Oct 26^th, 2010, 12:28pm

Sure, just use the one for -sin(x) at 0

Title: Re: power of cosine function
Post by pex on Oct 27^th, 2010, 12:31am

on 10/26/10 at 05:20:06, towr wrote:

If you want
[cos(c)]^{n^2} -> exp{- 1/2 (cn)^2}
then you must have
cos(c) -> exp{- 1/2 c^2}
Since c is constant, the expressions on both sides have to be equal, and thus c must be 0. (Which means it falls under the original case, since k=cn is constant for c=0.)

I don't think we can simply cancel the exponent n² when throwing infinities around: if you just want both sides to tend to zero, it is sufficient that c is not a multiple of pi. (But that's not something I would use the "->" notation for...)

Title: Re: power of cosine function
Post by towr on Oct 27^th, 2010, 12:49am

Well, they do both tend to zero, generally; but I figure we're interested in asymptotic behaviour, i.e. that one function approaches the other as n gets larger (rather than that they both approach a common limit).
So in that case, their quotient should tend to 1; and then it seems to me you can just cancel the n² factors, because they don't qualitatively change the asymptotic behaviour.
If a(x)^x/b(x)^x goes to 1 as x increases, then a(x)/b(x) must go to 1 as x increases (a necessary, but not sufficient condition). But in our case a(x) and b(x) are constants, and if they're not equal, then the expression goes to 0 or +/-infinity.

Title: Re: power of cosine function
Post by comehome1981 on Oct 27^th, 2010, 9:18pm

Thanks for all you who replied.

All answers are helpful. I think I got some idea now.